A converse theorem to the Cauchy's estimation

Question

This is a follow-up question to About the limit of the coefficient ratio for a power series over complex numbers.

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Cauchy's estimation in complex analysis is a consequence of the Cauchy's integral formula, which can be stated as follow:

Theorem. If $f$ is holomorphic in an open set $\Omega$ that contains a closed disc $D$ centered at $w$ with radius $R$, then for all $n$

$\displaystyle{ \frac{|f^{(n)}(w)|}{n!} \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$, where $C$ is the boundary circle of $D$.

Note that if $f$ has power series expansion $\sum b_n (z-w)^n$ inside $D$, then $b_n = |f^{(n)}(w)|/n!$.

Consider the following question: If $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc except for a pole at $z_0$ on the unit circle, and $f$ has the power series expansion $\sum a_n z^n$ in the open unit disc. By the answer of Sivaram to the previous post, the coefficients $a_n$ of the expansion of $f$ will not converge to zero. I'm wondering if this can be proved by something that is a (partial) converse to the Cauchy's estimation, and apply it backwards we can ensure that $a_n$ will not go to zero.

I would like to know whether the following version exists:

Assume $f$ is holomorphic in an open set $\Omega$ that contains a closed unit disc except at $z_0$ on the unit circle, and $f$ has power series expansion $\sum a_n z^n$ inside the open unit disc. If for all $n$ large enough and all $R < 1$ we have

$\displaystyle{ |a_n| \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$,

where $D$ is a closed disc centered at $0$ with radius $R$ and $C$ is the boundary circle of $D$, then $f$ is also holomorphic at $z_0$.

This version may not work at all, but any reasonable modifications which can be used to prove that $a_n$ will be bounded away from zero are fine.

Question. Is there a reasonable version to the converse of the Cauchy's estimation?

Answer 1

Unfortunately nothing like that holds. Consider $f(z) = \sqrt{1 - z}$. $f(z)$ is analytic on ${\mathbb C} - \{(x,0): x > 1\}$ and satisfies Cauchy's estimates on circles of radius $r < 1$ since the function is analytic on the unit disc. However, try as you may, you cannot extend $f(z)$ to a neighborhood of $1$. Note that Cauchy's estimates are even satisfied on the circle of radius $1$, by taking limits as $r$ goes to $1$ of the $r < 1$ Cauchy estimates.

Answer 2

It is not true. For example, $f(z)=\exp\left(\frac{z+1}{z-1}\right)$ is bounded by $1$ on the unit disk and all the coefficients are also bounded by $1$. But $f$ has an essential singularity at $z=1$.